Let $\mathrm{A}=\left\{\mathrm{x} \in(0, \pi)-\left\{\frac{\pi}{2}\right\}: \log _{(2 / \pi)}|\sin \mathrm{x}|+\log _{(2 / \pi)}|\cos \mathrm{x}|=2\right\}$ and $B=\{x \geq 0: \sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\}$. Then $\mathrm{n}(\mathrm{A} \cup \mathrm{B})$ is equal to:

Trigonometric Equations — JEE Main · 2025 · 24 Jan 2025 · Shift 2 Question with Solution

Let

A = {x \in (0, π) - {\frac{π}{2}} : lo g_{(2/ π)} ∣ sin x ∣ + lo g_{(2/ π)} ∣ cos x ∣ = 2}

and

B = {x \geq 0 : x (x - 4) - 3∣ x - 2∣ + 6 = 0}

. Then

n (A \cup B)

is equal to:

Answer: (C)

For set

A

lo g_{2/ π} ∣ sin x ∣ + lo g_{2/ π} ∣ cos x ∣ = 2

means

lo g_{2/ π} (∣ sin x cos x ∣) = 2

, i.e.

∣ sin x cos x ∣ = (\frac{2}{π})^{2} \Rightarrow \frac{∣ sin 2 x ∣}{2} = \frac{4}{π ^{2}} \Rightarrow ∣ sin 2 x ∣ = \frac{8}{π ^{2}} .

Since

0 < \frac{8}{π ^{2}} < 1

, this is solvable. For

x \in (0, π) - {\frac{π}{2}}

2 x

ranges over

(0, 2 π) - {π}

. The equation

∣ sin 2 x ∣ = k

for

0 < k < 1

has exactly 4 solutions in a full period

(0, 2 π)

(two where

sin 2 x = k > 0

and two where

sin 2 x = - k

), and none of these coincide with

2 x = π

(where

sin 2 x = 0 \neq = k

). So

∣ A ∣ = 4

.

For set

B

: let

t = x \geq 0

, so the equation is

t (t - 4) - 3∣ t - 2∣ + 6 = 0

.

For

t < 2

∣ t - 2∣ = 2 - t

, giving

t^{2} - 4 t - 3 (2 - t) + 6 = 0 \Rightarrow t^{2} - t = 0 \Rightarrow t = 0

t = 1

, both valid (i.e.

x = 0, 1

).

For

t \geq 2

∣ t - 2∣ = t - 2

, giving

t^{2} - 4 t - 3 (t - 2) + 6 = 0 \Rightarrow t^{2} - 7 t + 12 = 0 \Rightarrow t = 3

t = 4

, both valid (i.e.

x = 9, 16

).

So

B = {0, 1, 9, 16}

∣ B ∣ = 4

.

Since

A

consists of transcendental values of

x

(0, π)

(from

sin 2 x = \pm 8/ π^{2}

) that don't coincide with the integers

{0, 1, 9, 16}

(and

9, 16 > π

in any case),

A

and

B

are disjoint. So

n (A \cup B) = 4 + 4 = 8.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option